Trigonometry related to Kepler's Equation

solarblast

In deriving Kepler's Equation, there's a little trig work that gets at nu (see figure), which is the true anomaly. I'm curious how the third equation makes it into the 4th (last) equation. It's been a very long time since I've played with trig. nu of course is the atan of the last equation.

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DonAntonio

You should write down, with LaTeX, directly the equations here so that it'll be possible to check the notation and give, eventually, and answer, otherwise

it is way too cumbersome.

DonAntonio

solarblast

See <http://rip94550.wordpress.com/2011/05/02/elliptical-orbits-%E2%80%93-deriving-keplers-equation/>. Attached is the part I was looking for.

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D H

You're using the wrong equation as the source. Use the first two. From the second, compute [itex]1+\cos\nu[/itex]. Next compute [itex]sin\nu/(1+\cos\nu)[/itex]. This is just [itex]\tan(\nu/2)[/itex] per the half angle formula. Use the same formula to simplify the right hand side.

solarblast

What source are you referring to? The link or one of the equations in the attachment? The link is 10 pages long.

D H

I'm referring to the attachment in your first post.

solarblast

Ah, that was produced by a CM "pdf book" on the web. I'm satisfied with the longer attachment. I may contact the author of the pdf.